Let P and Q be the points of trisection of AB i.e., AP = PQ = QB
Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are
\(\left(\frac{1(-7)+2(2)}{1+2}, \frac{1(4)+2(-2)}{1+2}\right), \)i.e., (–1, 0)
Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
\(\left(\frac{2(-7)+1(2)}{2+1}, \frac{2(4)+1(-2)}{2+1}\right)\) ,i.e., (– 4, 2)
Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

From the given figure, the position of green flag posted by Niharika is M \(\left(2 \times 1, \frac{1}{4} \times 100\right)\) i.e. M (2, 25) and red flag posted by Preet is N\(\left(8 \times 1, \frac{1}{5} \times 100\right)\) i.e. N(8, 20).
Now, \(\begin{aligned} & M N= \sqrt{(8-2)^2+(20-25)^2} \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right] } \\ \end{aligned}\)
\(\begin{aligned} &=\sqrt{(6)^2+(-5)^2}=\sqrt{36+25}=\sqrt{61} \end{aligned}\)
Hence, the distance between flags is \(\sqrt{61} \mathrm{~m}\)
Let P be the position of the blue flag posted by Raimi in the halfway of line segment MN.
Then, coordinates of P = \(\begin{aligned} \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\ \end{aligned}\)
\(=\left(\frac{2+8}{2}, \frac{25+20}{2}\right)=\left(\frac{10}{2}, \frac{45}{2}\right)=(5,22.5)\)
Hence, the blue flag is on the 5th line at a distance of 22.5 m above it.

From the given figure,the coordinates of points A, B, C and D are (3,4), (6, 7), (9, 4) and (6, 1),respectively.
Now, \(\begin{aligned} A B=\sqrt{(6-3)^2+(7-4)^2} \end{aligned}\)
\(\begin{aligned} \left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right] \end{aligned}\)
\(\begin{aligned} & =\sqrt{3^2+3^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \text { units } \\ \end{aligned}\)
\(\begin{aligned} BC =\sqrt{(9-6)^2+(4-7)^2} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{3^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \text { units } \\ \end{aligned}\)
\(\begin{aligned} CD =\sqrt{(6-9)^2+(1-4)^2} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \text { units } \end{aligned}\)
\(\begin{aligned} DA=\sqrt{(3-6)^2+(4-1)^2} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{(-3)^2+3^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \text { units } \\ \end{aligned}\)
\(\begin{aligned} AC=\sqrt{(9-3)^2+(4-4)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{36+0}=\sqrt{36}=6 \text { units } \\ \end{aligned}\)
and \(\begin{aligned} BD=\sqrt{(6-6)^2+(1-7)^2}=\sqrt{0+36}=\sqrt{36}=6 \text { units } \end{aligned}\)
Here, all the four sides (AB, BC, CD, DA) are equal and diagonals (AC, BD) are equal.
Hence, ABCD is a square. So, Champa is correct.

Let A(2, 3) and B(4, 1) be the given points.
Here, x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1
Now, AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
[by distance formula]
\(\begin{aligned} & =\sqrt{(4-2)^2+(1-3)^2} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \end{aligned}\)

Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

A is a point on x-axis.
∴ 3x+4x0=24 ⇒ x=8
⇒ OA=8 ⇒ OB=6
Similarly, coordinates of B are (0,6)
Area of △AOB = \(\frac{1}{2}\)xOAxOB
=  \(\frac{1}{2}\)x8x6=24 sq. units

\(a=2,\ OR\ \left( \triangle ABC \right) =6\ sq.units\)

△ABC is an equilateral triangle with  AB = BC = AC. 
Let the coordinates of C be (x, y) 
⇒ AC=AB ⇒ AC²=AB²
⇒ (x-3)²+(y-4)²=(-2-3)²+(3-4)²
⇒ x²+9-6y+y²+16-8y=(-5)²+(-1)²
⇒ x²+y²-6x--8y+25=25+1
⇒ x²+y²-6x--8y=26-25
⇒ x²+y²-6x--8y=1 ---- (i)
∴  AC=BC
∴  AC²=BC²
⇒ (x-3)²+(y-4)²=(x+2)²+(y-3)²
⇒ x²+9-6x+y²+16-8y
=x²+4+4x+y²+9-6y
⇒ 25-6x-8y=13+4x-6y
⇒ 4x+6x-6y+8y=25-13
⇒ 10x+2y=12 ⇒ 5x+y=6
⇒ y=6-5x    --- (ii)
Substituting this value of y from (ii) in (i), we get
x²+(6-5x)²-6x-8(6-5x)=1
⇒ x²+36+25x²-60x-60x-48+40x=1
⇒ 26x²-26x-12-1=0
⇒ 26x²-26x-13=0
⇒ 2x²-2x-1=0
\(\Rightarrow \ x=\frac { -(-2)\pm \sqrt { ({ -2) }^{ 2 }-4\times 2\times (-1) } }{ 4 } \)
\(\ =\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Putting this value of x in equation (ii ), we get
\(y=6-5\left( \frac { 1\pm \sqrt { 3 } }{ 2 } \right) =\frac { 12-5\left( 1\pm \sqrt { 3 } \right) }{ 2 } \)
\(=\frac { 12-5\pm 5\sqrt { 3 } }{ 2 } =\frac { 7\pm 5\sqrt { 3 } }{ 2 } \)
Thus, coordinates of the third vertex are
\(\left( \frac { 1\pm \sqrt { 3 } }{ 2 } ,\frac { 7\pm 5\sqrt { 3 } }{ 2 } \right) \)

A.T.Q. Diameter = 10\(\sqrt {2}\)
OA²=50
⇒ (2a-11)²+(a-7+9)²=50
⇒​ 4a²+121-44a+(a+2)²=50
⇒ 4a²-44a+121+a²+4+4a-50=0
⇒ 5a²-40a+75=0,
⇒ a²-8a+15=0
⇒ (a-5)(a-3)=0 ⇒ a=5,3​​​​​​

Using distance formula
\(AB=\sqrt { ({ -3-2) }^{ 2 }+[8-({ -2)] }^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+10^{ 2 } } =\sqrt { 25+100 } \)
\(=\sqrt { 125 } =5\sqrt { 5 } units\)
\(BC=\sqrt { [{ -1-(-3)] }^{ 2 }+(4-8)^{ 2 } } \)
\(=\sqrt { { 2 }^{ 2 }+4^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ -1-2) }^{ 2 }+[4-(-2)]^{ 2 } } \)
\(=\sqrt { { 3 }^{ 2 }+6^{ 2 } } =\sqrt { 9+36 } =\sqrt { 45 } \)
\(=3\sqrt { 5 } units\)
\(\because \ 2\sqrt { 5 } +3\sqrt { 5 } =5\sqrt { 5 } \)
∴ Points are collinear. 

Let point on y-axis be (0,a)
Now distance of this point from (5,-2) equal to distance from point (-3,2)
i.e.,  \(\Rightarrow \sqrt { 5^{ 2 }+\left( -2-a \right) ^{ 2 } } =\sqrt { \left( 3 \right) ^{ 2 }+\left( a-2 \right) ^{ 2 } } \)
Squaring and simplifying, we get
25+4+a²+4a = 9+a²+4-4a ⇒ 8a=-16  ⇒ a=-2

Since the point P(x,y) is equidistant from the points A(7, 1) and B(3, 5). 
Therefore PA = PB ...(i)
Using distance formula
\(PA=\sqrt { \left( x-7 \right) ^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(=\sqrt { { x }^{ 2 }+{ 7 }^{ 2 }-2.7x+{ y }^{ 2 }+{ 1 }^{ 2 }-2.y.1 } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+49+1-14x-2y } \)
\(PB=\sqrt { \left( x-3 \right) ^{ 2 }+\left( y-5 \right) ^{ 2 } } \)  (using distance formula)
\(=\sqrt { { x }^{ 2 }+{ 3 }^{ 2 }-2.3x+{ y }^{ 2 }+{ 5 }^{ 2 }-2.5.y } =\sqrt { { x }^{ 2 }+9-6x+{ y }^{ 2 }+25-10y } \)
Substituting the values of PA and PB in (i). we get
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 }+50-14x-2y } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }-6x-10y+34 } \)
Squaring both sides, We get
\({ x }^{ 2 }+{ y }^{ 2 }+50-14x-2y={ x }^{ 2 }+{ y }^{ 2 }-6x-10y+34\)
⇒ 50-34=14x+2y-6x-10y ⇒16=8x-8y ⇒x-y=2

Now, AP:PB = 1:2
\(p={1\times1+2\times3\over 1+2}\Rightarrow={7\over 3}\)
Also AQ:QB = 2:1\(\Rightarrow q={2\times2+1\times-4\over 1+2}=0\)

AB = \(\sqrt { ({ 2-7) }^{ 2 }+{ (3-5) }^{ 2 } } =\sqrt { 25+4 } =\sqrt { 29 } \)
BC = \(\sqrt { ({ 6-2) }^{ 2 }+{ (-7-3) }^{ 2 } } =\sqrt { 16+100 } =\sqrt { 116 } \)
CA = \( \sqrt { ({ 7-6) }^{ 2 }+{ (5+7) }^{ 2 } } =\sqrt { 1+144 } =\sqrt { 145 } \)
Since AB²+BC² = 29+116 = 145 =CA².
∴ △ABC is right angled at B.
Area = \(\frac{1}{2} AB \times BC = \frac {1}{2}=\sqrt{29}.\sqrt{116}=\frac {1}{2}\sqrt{29.2}.{2}\sqrt{29}=29\)

A(1,2), B(5,4), C(3,8) and D(-1, 6)
\(AB=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20};\ BC=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
\(CD=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20};\ DA=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
Here AB=BC=CA=DA
\(AC=\sqrt{2^2+6^2}=\sqrt{40}\ and\ BD=\sqrt{(-6)^2+(2)^2}=\sqrt{36+4}=\sqrt{40}\)
All sides of quadrilateral are equal and diagonals are equal.
ABCD is square.